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2026 VCE Chemistry exam: What to learn from last year's mistakes
Hi all, in preparation for the upcoming exam, here's what to learn from last year's mistakes.
2025 median exam grade: C+. VCAA does not publish an exact median mark.
Average: 61.05/120, or 50.9%. VCAA reports 122.1/240; this is converted to the exam's 120-mark scale.
10,873 students assessed. 8.7% received A+. The 2025 A+ cutoff was 187/240, equivalent to 93.5/120. That is last year's cutoff, not a prediction for 2026.
Figures: VCAA's 2025 grade distribution, written examination on page 3: source.
Question averages come from the 2025 examiner report, updated 21 July 2026: source. Percentages are rounded. My examples illustrate the fixes; they are not full model answers. All questions below are from Section B.
2e: Comparing electroplating currents
Average: 0.6/2. 69% received zero. 23% received full marks.
Where students went wrong: They compared electrode potentials instead of working out how many electrons were needed. Some missed that both metal ions had the same charge.
What they should have done: Connect molar mass to moles of metal, then electrons, then current.
For the same 7.8 g deposit, rhodium needs fewer moles because its molar mass is larger. Both ions require three electrons per ion, so rhodium also needs fewer moles of electrons. For the same 2.5 hours, that means a smaller current: about 2.44 A for rhodium compared with 4.83 A for chromium.
4d: Using the proton NMR evidence
Average: 0.8/3. 46% received zero. Only 5% received full marks.
Where students went wrong: They stated features of the molecule without tying each one to the supplied spectrum or table.
What they should have done: Pair each structural claim with the observation supporting it.
Three signals indicate three hydrogen environments. All three are singlets, so those environments have no adjacent hydrogens producing splitting. The areas 16.7:33.3:100 simplify to 1:2:6. Those are three separate evidence-based deductions. A list of possible functional groups does not do the same job.
4a(ii): Giving the reactant names requested
Average: 0.6/2. 56% received zero. Only 17% received full marks.
Where students went wrong: A high proportion supplied formulas when the question asked for names.
What they should have done: Name both reactants for the substitution reaction producing propan-1-amine.
One accepted pair was 1-chloropropane and ammonia. Writing C₃H₇Cl and NH₃ does not follow the instruction to name them. Before moving on, check whether the question wants a name, formula, equation or structure. Here, that distinction affected a two-mark response.
6e: Scaling the titration result back to the spinach sample
Average: 1.7/4. 36% received zero. 13% received full marks.
Where students went wrong: They missed the reaction ratio or the dilution factor.
What they should have done: Keep track of what sample each number describes.
The titre contains 3.084 × 10⁻⁵ mol of permanganate. Multiply by 5/2 to get 7.71 × 10⁻⁵ mol of oxalate in the 25 mL aliquot. Then multiply by 500/25 to recover the amount in the full solution: 1.542 × 10⁻³ mol. Multiplying by 88.0 gives 0.136 g, reported as 0.14 g. The aliquot is only one twentieth of the solution.
4b(i): Explaining boiling points
Average: 1.0/2. 35% received zero. 29% received full marks.
Where students went wrong: They discussed the strength of covalent bonds within the molecules, or claimed amines only have dispersion forces.
What they should have done: Compare attractions between molecules, then explain the energy needed to separate them.
The carboxylic acids can form hydrogen-bonded dimers. Stronger attraction between their molecules means more energy is needed to separate them during boiling. You are not breaking the C=O bonds inside each molecule when the liquid boils.
3b(ii): Pressure and equilibrium
Average: 1.2/2. 23% received zero. 37% received full marks.
Where students went wrong: They said increasing pressure favours the forward reaction without explaining how the pressure increased.
What they should have done: Specify a decrease in volume and use the gas-particle ratio.
For CO + 2H₂ ⇌ CH₃OH, decreasing the volume increases pressure and favours the side with fewer gas particles. That is the product side: one compared with three. Adding an inert gas at constant volume also raises total pressure, but does not produce that same equilibrium shift.
3d: Initial amount of hydrogen
Average: 1.9/4. 33% received zero. 28% received full marks.
Where students went wrong: They did not combine the equilibrium expression with the changes in concentration. Some finished with a concentration when the question asked for moles.
What they should have done: Find the equilibrium hydrogen concentration, then recover the initial amount using the reaction ratio and container volume.
Forming 1.30 M of methanol consumes 1.30 M of CO and 2.60 M of H₂. The equilibrium CO concentration is therefore 3.70 M. Using K gives an equilibrium H₂ concentration of about 0.539 M. Add back the 2.60 M consumed, then multiply by the 1.00 L volume: initially 3.14 mol of H₂. Do not stop at the equilibrium amount.
Try B2e, B4d and B6e before reading their report sections. Together they are worth 9 marks. Label every quantity in the calculation and give the evidence behind each written claim.
Use the VCE Chemistry hub and past papers and examiner reports alongside this.
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